Question: Solve for $r$, $ \dfrac{8}{5r + 1} = -\dfrac{4}{20r + 4} + \dfrac{r}{15r + 3} $
Answer: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5r + 1$ $20r + 4$ and $15r + 3$ The common denominator is $60r + 12$ To get $60r + 12$ in the denominator of the first term, multiply it by $\frac{12}{12}$ $ \dfrac{8}{5r + 1} \times \dfrac{12}{12} = \dfrac{96}{60r + 12} $ To get $60r + 12$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{4}{20r + 4} \times \dfrac{3}{3} = -\dfrac{12}{60r + 12} $ To get $60r + 12$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{r}{15r + 3} \times \dfrac{4}{4} = \dfrac{4r}{60r + 12} $ This give us: $ \dfrac{96}{60r + 12} = -\dfrac{12}{60r + 12} + \dfrac{4r}{60r + 12} $ If we multiply both sides of the equation by $60r + 12$ , we get: $ 96 = -12 + 4r$ $ 96 = 4r - 12$ $ 108 = 4r $ $ r = 27$